\newproblem{lay:5_3_27}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.3.27}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Show that if $A$ is both diagonalizable and invertible, then so is $A^{-1}$
}{
   % Solution
	If $A$ is diagonalizable, then there exist an invertible matrix $P$ and a diagonal matrix $D$ such that
	\begin{center}
		$A=PDP^{-1}$
	\end{center}
	If $A$ is invertible, then
	\begin{center}
		$A^{-1}=(PDP^{-1})^{-1}=(P^{-1})^{-1}D^{-1}P^{-1}=PD^{-1}P^{-1}$
	\end{center}
	So, $D$ is also invertible and we see that $A^{-1}$ is also diagonalizable.
}
\useproblem{lay:5_3_27}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
